題解

感覺智商為0啊QAQ

顯然對於一個長度為\(len\)的border，每個點同餘\(n - len\)的部分必然相等

那麼我們求一個\(f[a]\)陣列，如果存在\(s[x] = 0\)且\(s[y] = 1\)且\(|x - y| = a\)

這個很好求，只要把0和1分別挑出來，NTT卷一下就好了

一個\(len\)合法，即它的\(n - len\)的倍數\(k\)，\(f[k]\)都等於0

程式碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 500005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353,MAXL = (1 << 20);
int W[MAXL + 5],f[MAXL + 5],g[MAXL + 5],N;
char s[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
void NTT(int *p,int len,int on) {
    for(int i = 1 , j = len >> 1 ; i < len - 1 ; ++i) {
    if(i < j) swap(p[i],p[j]);
    int k = (len >> 1);
    while(j >= k) {
        j -= k;
        k >>= 1;
    }
    j += k;
    }
    for(int h = 2 ; h <= len ; h <<= 1) {
    int wn = W[(MAXL + MAXL / h * on) % MAXL];
    for(int k = 0 ; k < len ; k += h) {
        int w = 1;
        for(int j = k ; j < k + h / 2 ; ++j) {
        int u = p[j],t = mul(p[j + h / 2],w);
        p[j] = inc(u,t);
        p[j + h / 2] = inc(u,MOD - t);
        w = mul(w,wn);
        }
    }
    }
    if(on == -1) {
    int InvL = fpow(len,MOD - 2);
    for(int i = 0 ; i < len ; ++i) p[i] = mul(p[i],InvL);
    }
}
void Init() {
    W[0] = 1;W[1] = fpow(3,(MOD - 1) / MAXL);
    for(int i = 2 ; i < MAXL ; ++i) {
    W[i] = mul(W[i - 1],W[1]);
    }
    scanf("%s",s + 1);
}
void Solve() {
    int t = 1;
    N = strlen(s + 1);
    while(t <= 2 * N) t <<= 1;
    for(int i = 1 ; i <= N ; ++i) {
    f[i] = (s[i] == '1');
    g[i] = (s[N - i + 1] == '0');
    }
    NTT(f,t,1);NTT(g,t,1);
    for(int i = 0 ; i < t; ++i) f[i] = mul(f[i],g[i]);
    NTT(f,t,-1);
    int64 ans = 1LL * N * N;
    for(int i = 1 ; i < N ; ++i) {
    int t = i;
    bool flag = 0;
    while(t < N) {
        if(f[N - t + 1] || f[N + t + 1]) {flag = 1;break;}
        t += i;
    }
    if(!flag) ans ^= 1LL * (N - i) * (N - i);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}